1. Triangle ABC has side lengths AB=9, AC=10, and BC=17. Let X be the intersection of the angle bisector of angle A with side \(\overline{BC}\), and let Y be the foot of the perpendicular from X to side \(\overline{AC}\). Compute the length of \(\overline{XY} \)
2. \(\overline{BC}\) is a chord of a circle with center O and area \(48\pi \). Point A is inside \(\triangle BCO\) such that \(\triangle ABC\) is equilateral and A is the circumcenter of \(\triangle BCO\). What is the area of triangle ABC?
Thank you!!!
First one....see the image below :
Let A = (0,0)
Let C = (10,0)
And we can find B, thusly
Construct a circle with radius 9 centered at the origin
The equation of this circle will be x^2 + y^2 = 81 (1)
And construct a circle wilth radius 17 centered at (10, 0)
The equation of this circle will be (x - 10)^2 + y^2 = 289 (2)
We can find the coordinates for point B, thusly :
Subtract (1) from (2) and we have that
(x-10)^2 - x^2 = 208 simplify
x^2 - 20x + 100 - x^2 = 208
-20x = 108
x = -108/20 = -5.4 = x coordinate of B
And taking the positive value for y, we have that
(-5.4)^2 + y^2 = 81
y^2 = 81 - 29.16
y = 7.2 = y coordinate of B
So....B = (-5.4, 7.2)
And AX is the bisector of angle A
So....we have the following relationship
AC/AB = XC /XB
10/9 = XC / XB
Then BC has 19 equal parts and XC is 10 of them.....so XC = (10/19)*17 = 170/19
So....using a similar triangle idea.....
Distance from B to the x axis / BC = XY / XC
7.2 / 17 = XY / (170/19)
(170/19) * 7.2 /17 = XY = (170/17)(7.2 / 19) = 10 * 7.2 / 19 = 72 / 19
Second one.....see the following image....
Let O = (0,0)
Since the area of circle O = 48pi....then the radius = √48 units
So...the equation of the circle is x^2 + y^2 = 48
Since ABC is equilateral....then angle BAC = 60°
Then the excluded angle to BAC = 300°
So, by symmetry, angle AOC = 150°
And since A is the circumcenter of triangle BOC, then AO = AC = the side of equilateral triangle ABC = s
And we can find s^2 using the Law of Cosines, thusly :
OC^2 = 2s^2 - 2s^2 * ( cos 150)
48 = 2s^2 + 2s^2* (√3/2)
48 = 2s^2 + s^2√3
48 = (2 + √3)s^2
48 / [ 2 + √3 ) = s^2 = 48 [ 2 - √3] = 96 - 48√3
So....the area of equilateral triangle ABC =
(1/2) s^2 sin (BAC) =
(1/2)s^2 sin(60°) =
(√3) [ 96 - 48√3 ] / 4 =
(√3) [ 24 - 12√3] =
24√3 - 36 units^2